How far will it go?

**CivilWars** · 01-31-08, 02:18 PM

I was inspired by the will it fly also. This debate has been going on in my family for generations, so maybe we can either solve it, or argue it more, here.

Suppose you are on a train heading East. You are standing on the back of the train facing West. You hav a gun, and for arguments sake the gun fires its bullet at a rate of X MPH. The train is also traveling at X MPH.

If you fire the gun at the moment the train passes point A where will the bullet land in relation to point A?

**rv2pc3d** · 01-31-08, 03:10 PM

here's my $.002:

I believe the bullets inertia is being affected by the energy that was released from the ignition of the gun powder in the shell. So I believe the bullet will continue to move forward (westward, past point A) until affected by an opposing force; that force being Gravity and friction, vapor etc. The question then becomes, at what distance/side beyond Point A will it fall.

Assuming POINT A is a stationary point on a graph and that the earth is rotating independently of that point, then the distance would be determined by calculating the speed of the earth's rotation, in conjunction with gravity and friction with the speed of the train (X). In which case, the bullet should be landing to the west of POINT A.

Assuming POINT A moves with the rotation of the earth, then the bullet will land to the EAST of PIONT A, assuming the Earth is moving at a speed greater than the train (X).

Even if the bullet were dropped without any energy to have propelled it, I think it will still land to the EAST of POINT A. Some Inertial energy from you and the train should be transfered to the object upon the moment of release. Once the object has been released, gravity will take over inertia and begin in straight deceleration, unless acted upon by an opposing force; i.e., distance, wind, vapor, etc.

I'm no scientist... but glad I could help with your "homework"...! LOL! (assuming I'm correct, if not, then theres always a job for you at Wendy's).

-modified for spelling

**rv2pc3d** · 01-31-08, 03:21 PM

Option #2:

Assuming I would be from the West, heading East on a train... then the bullet might be lodged in my brain, in which case, I would fall off the cabuse, landing dead center on POINT A.

Thats sounds logical.

**rv2pc3d** · 01-31-08, 03:25 PM

Come on, NO takers.... I see several votes but no explanations.

I guess everyones on the ol' INTERNET looking it up. I should do the same...

**Mcstrange** · 01-31-08, 03:35 PM

The whole Newtonian laws and what not.
Object in motion (bullet) traveling at 1000 Feet Per Second as it is in a gun on a train; is acted upon by a force (being fired) that causes it to move at that same rate 1000fps in the opposite direction.
It will drop to the ground and land in a spot very close relatively to the point at which it was fired.

**rv2pc3d** · 01-31-08, 03:44 PM

Originally Posted by Mcstrange

The whole Newtonian laws and what not.
Object in motion (bullet) traveling at 1000 Feet Per Second as it is in a gun on a train; is acted upon by a force (being fired) that causes it to move at that same rate 1000fps in the opposite direction.
It will drop to the ground and land in a spot very close relatively to the point at which it was fired.

Close Yes, but shy of that point I think

?; as the POINT continues to move without the affects of the opposing forces away fromthe bullet.

**Veovis** · 01-31-08, 03:50 PM

Originally Posted by Mcstrange

The whole Newtonian laws and what not.

Imagine that.

**DJ Ms. White** · 01-31-08, 03:56 PM

Without taking the rotation of the earth or anything such as friction into consideration (therefore assuming ideal circumstances), the bullet will stay in the same place. The magnitude of the velocity of the bullet is the speed of the bullet relative to the train, -X, plus the speed of the train, X. (This is assuming east to be the positive direction.) Thus the bullet stays in the same place.

Taking air friction into consideration, the air friction on the train is irrelevant as the speed of the train at the time the bullet is fired is X. This value does not change at the point in time. At other times yes it will vary, but from the stand point of this problem that does not matter. The bullet's speed will however decrease due to the acceleration acting on it caused by the force of the air friction.
Applying the same equation Vbullet=X-(X-africtiont) the bullets velocity will be positive and to the east (This assumes a constant deceleration due to the air friction.)

This is also what occurs from the frame of reference of the Earth. As the point A, the bullet, and the train all experience the same velocity of the frame of reference of the Earth.
From the point of view of something away from Earth assuming the point A to exist on a spherical plane that covers the Earth's surface yet is separate from it, yes the position will vary. However, that would require the use of some nasty linear algebra I don't know along with the general theory of relativity.
---edit note the above equation, Vbullet=X-(X-africtiont) , reduces to Vbullet=africtiont

**rv2pc3d** · 01-31-08, 04:19 PM

Originally Posted by DJ Mr. White

Without taking the rotation of the earth or anything such as friction into consideration (therefore assuming ideal circumstances), the bullet will stay in the same place. The magnitude of the velocity of the bullet is the speed of the bullet relative to the train, -X, plus the speed of the train, X. (This is assuming east to be the positive direction.) Thus the bullet stays in the same place. {There then would be no bullet propulsion if we discount "anything such as friction" ,etc.... the whole point becomes MUTE if we don't include these forces.}

**DJ Ms. White** · 01-31-08, 04:42 PM

We know that the bullet leaves the gun with a velocity X, so the friction forces exerted on the bullet by the barrel are irrelevant in terms of solving this problem. Furthermore, the bullet would still travel in a vacuum as when it left the gun, it had momentum and by the laws of physics, momentum must be conserved. Remember, that the problem has essentially done the calculations of what happened inside of the gun to determine the muzzle velocity of the bullet and the summation of forces acting on the train to give us the train's velocity at the point in time the gun was fired.

Furthermore, as I stated above the rotation of the Earth does not affect the bullets position from the standpoint of the Earth. You could get nit picky and say that the gravitational acceleration on the bullet and the train will be different based upon the vertical position of each, but the difference a few meters makes with the low amounts of mass in this solar system is inconsequential. Now if the bullet was in orbit around the Earth say a few hundred thousand miles up, yes there would be a difference, but this is not the case.